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        <h1 id="题目链接"><a href="#题目链接" class="headerlink" title="题目链接"></a><a href="http://poj.org/problem?id=3728" target="_blank" rel="noopener">题目链接</a></h1><h2 id="题意："><a href="#题意：" class="headerlink" title="题意："></a>题意：</h2><p>给定一个N个节点的树，1&lt;=N&lt;=50000 每个节点都有一个权值，代表商品在这个节点的价格。商人从某个节点a移动到节点b，且只能购买并出售一次商品，问最多可以产生多大的利润。Q次询问<a id="more"></a></p>
<h2 id="思想与解法"><a href="#思想与解法" class="headerlink" title="思想与解法"></a>思想与解法</h2><ol>
<li>首先最暴力的想法就是去找到路径然后判断，找路径的办法就是先找到$LCA$然后暴力就完事了，很明显时间复杂度$O(N*Q)$绝对$tle$</li>
<li>我们思考得到最大利润的方式，首先对于$u \rightarrow v$可以分解为$u \rightarrow LCA \rightarrow v$<br>那么最大利润将会如下三条路径中产生：<br>1.$u \rightarrow LCA$<br>2.$LCA \rightarrow v$<br>3.$u \rightarrow v$<br>这时候可能有很多人会想为什么要把路径3分为路径1与路径2，明显路径3包含1和2啊。<blockquote>
<p>在这里解释一下，如果只观察路径3的话，那么答案就是路径上最大的权值减最小权值，但是，无法记录访问顺序！但是这样分开以后以LCA为参考，获得每个点到$LCA$路径的最大利润，如果$u \rightarrow v$的最大利润不在点到$LCA$的路径而是穿过路径的时，那么商人一定在$u \rightarrow LCA$进货，在$LCA \rightarrow v$出售。直接记录两个路径中的最大最小值然后计算就好了</p>
<h2 id="操作"><a href="#操作" class="headerlink" title="操作"></a>操作</h2><p>根据分析我们需要四个数组！一个储存路径1的最大利润$up$，一个储存路径2的最大利润$down$，第三个储存路径的最大值$ma$，第四个存储最小值$mi$。然后并查集，把每一个点并到$LCA$上并更新所有数据。最后统计答案，据说卡vector。。<br>并查集的合并操作</p>
</blockquote>
</li>
</ol>
<figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br></pre></td><td class="code"><pre><span class="line">int get(int x)&#123;</span><br><span class="line">    if(fa[x]==x) return x;</span><br><span class="line">    int root = fa[x];</span><br><span class="line">    fa[x]=get(fa[x]);</span><br><span class="line">    up[x]=max(up[x],max(up[root],ma[root]-mi[x]));</span><br><span class="line">    down[x]=max(down[x],max(down[root],ma[x]-mi[root]));</span><br><span class="line">    mi[x]=min(mi[x],mi[root]);</span><br><span class="line">    ma[x]=max(ma[x],ma[root]);</span><br><span class="line">    return fa[x];</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure>
<p>卡vector我们就是用链式前向星寸询问，继而离线！$add1$函数<br>由于我们要在合并完以后才能得到答案，所以我的办法是每一次计算完$LCA$，用链式前向星保存在$LCA上(add2)！$访问到时在计算ans。</p>
<h2 id="代码"><a href="#代码" class="headerlink" title="代码"></a>代码</h2><figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br><span class="line">41</span><br><span class="line">42</span><br><span class="line">43</span><br><span class="line">44</span><br><span class="line">45</span><br><span class="line">46</span><br><span class="line">47</span><br><span class="line">48</span><br><span class="line">49</span><br><span class="line">50</span><br><span class="line">51</span><br><span class="line">52</span><br><span class="line">53</span><br><span class="line">54</span><br><span class="line">55</span><br><span class="line">56</span><br><span class="line">57</span><br><span class="line">58</span><br><span class="line">59</span><br><span class="line">60</span><br><span class="line">61</span><br><span class="line">62</span><br><span class="line">63</span><br><span class="line">64</span><br><span class="line">65</span><br><span class="line">66</span><br><span class="line">67</span><br><span class="line">68</span><br><span class="line">69</span><br><span class="line">70</span><br><span class="line">71</span><br><span class="line">72</span><br><span class="line">73</span><br><span class="line">74</span><br><span class="line">75</span><br><span class="line">76</span><br><span class="line">77</span><br><span class="line">78</span><br><span class="line">79</span><br><span class="line">80</span><br><span class="line">81</span><br><span class="line">82</span><br><span class="line">83</span><br><span class="line">84</span><br><span class="line">85</span><br><span class="line">86</span><br><span class="line">87</span><br><span class="line">88</span><br><span class="line">89</span><br><span class="line">90</span><br><span class="line">91</span><br><span class="line">92</span><br><span class="line">93</span><br><span class="line">94</span><br><span class="line">95</span><br><span class="line">96</span><br><span class="line">97</span><br><span class="line">98</span><br><span class="line">99</span><br><span class="line">100</span><br><span class="line">101</span><br><span class="line">102</span><br></pre></td><td class="code"><pre><span class="line">#include&lt;stdio.h&gt;</span><br><span class="line">#include&lt;algorithm&gt;</span><br><span class="line">#include&lt;string.h&gt;</span><br><span class="line">using namespace std;</span><br><span class="line">const int N = 3e5+5;</span><br><span class="line">struct ED&#123;</span><br><span class="line">    int pre,id;</span><br><span class="line">&#125;ed[N],ed1[N];</span><br><span class="line">int head[N],head1[N],tot,tot1,ans[N],fa[N],ma[N],mi[N],vis[N],up[N],down[N],head2[N],tot2=1;</span><br><span class="line">struct NUM&#123;</span><br><span class="line">    int x,y,id,pre;</span><br><span class="line">&#125;num[N];</span><br><span class="line">void add(int u,int v)&#123;</span><br><span class="line">    ed[++tot].pre=head[u];</span><br><span class="line">    ed[tot].id=v;</span><br><span class="line">    head[u]=tot;</span><br><span class="line">&#125;</span><br><span class="line">void add1(int u,int v)&#123;</span><br><span class="line">    ed1[++tot1].pre=head1[u];</span><br><span class="line">    ed1[tot1].id=v;</span><br><span class="line">    head1[u]=tot1;</span><br><span class="line">&#125;</span><br><span class="line">void add2(int u,int x,int y,int id)&#123;</span><br><span class="line">    num[++tot2].id=id;</span><br><span class="line">    num[tot2].x=x,num[tot2].y=y;</span><br><span class="line">    num[tot2].pre=head2[u];</span><br><span class="line">    head2[u]=tot2;</span><br><span class="line">&#125;</span><br><span class="line">int get(int x)&#123;</span><br><span class="line">    if(fa[x]==x) return x;</span><br><span class="line">    int root = fa[x];</span><br><span class="line">    fa[x]=get(fa[x]);</span><br><span class="line">    up[x]=max(up[x],max(up[root],ma[root]-mi[x]));</span><br><span class="line">    down[x]=max(down[x],max(down[root],ma[x]-mi[root]));</span><br><span class="line">    mi[x]=min(mi[x],mi[root]);</span><br><span class="line">    ma[x]=max(ma[x],ma[root]);</span><br><span class="line">    return fa[x];</span><br><span class="line">&#125;</span><br><span class="line">void LCA(int x)&#123;</span><br><span class="line">    vis[x]=1;</span><br><span class="line">    int i;</span><br><span class="line">    for(i=head[x];~i;i=ed[i].pre)&#123;</span><br><span class="line">        int v=ed[i].id;</span><br><span class="line">        if(vis[v]) continue;</span><br><span class="line">        else &#123;</span><br><span class="line">            LCA(v);</span><br><span class="line">            fa[v]=x;</span><br><span class="line">        &#125;</span><br><span class="line">    &#125;</span><br><span class="line">    for(i=head1[x];~i;i=ed1[i].pre)&#123;</span><br><span class="line">        int v=ed1[i].id;</span><br><span class="line">        if(vis[v]) &#123;</span><br><span class="line">            add2(get(v),x,v,i);</span><br><span class="line">        &#125;</span><br><span class="line">    &#125;</span><br><span class="line">    for(i=head2[x];~i;i=num[i].pre)&#123;</span><br><span class="line">        int u = num[i].x,v=num[i].y,d=num[i].id;</span><br><span class="line">        get(u),get(v);</span><br><span class="line">        if(d%2)&#123;</span><br><span class="line">            ans[d]=ans[d^1]=max(up[v],max(down[u],ma[u]-mi[v]));</span><br><span class="line">        &#125;</span><br><span class="line">        else&#123;</span><br><span class="line">            ans[d]=ans[d^1]=max(down[v],max(up[u],ma[v]-mi[u]));</span><br><span class="line">        &#125;</span><br><span class="line">    &#125;</span><br><span class="line">&#125;</span><br><span class="line">void init()&#123;</span><br><span class="line">    memset(head,-1,sizeof head);</span><br><span class="line">    memset(head1,-1,sizeof head1);</span><br><span class="line">    memset(head2,-1,sizeof head2);</span><br><span class="line">    memset(vis,0,sizeof vis);</span><br><span class="line">    tot=tot1=tot2=1;</span><br><span class="line">&#125;</span><br><span class="line">int main()&#123;</span><br><span class="line">    int n,m,i,j,k;</span><br><span class="line">    while(scanf(&quot;%d&quot;,&amp;n)==1)&#123;</span><br><span class="line">        int u,v,w;</span><br><span class="line">        init();</span><br><span class="line">        for(i=1;i&lt;=n;i++) fa[i]=i;</span><br><span class="line">        for(i=1;i&lt;=n;i++)&#123;</span><br><span class="line">            scanf(&quot;%d&quot;,&amp;w);</span><br><span class="line">            down[i]=up[i]=0;</span><br><span class="line">            ma[i]=mi[i]=w;</span><br><span class="line">        &#125;</span><br><span class="line">        for(i=1;i&lt;n;i++)&#123;</span><br><span class="line">            scanf(&quot;%d %d&quot;,&amp;u,&amp;v);</span><br><span class="line">            add(u,v);</span><br><span class="line">            add(v,u);</span><br><span class="line">        &#125;</span><br><span class="line">        int q;</span><br><span class="line">        scanf(&quot;%d&quot;,&amp;q);</span><br><span class="line">        for(i=1;i&lt;=q;i++)&#123;</span><br><span class="line">            scanf(&quot;%d %d&quot;,&amp;u,&amp;v);</span><br><span class="line">            add1(u,v);</span><br><span class="line">            add1(v,u);</span><br><span class="line">        &#125;</span><br><span class="line">        LCA(1);</span><br><span class="line">        for(i=2;i&lt;=2*q;i+=2)&#123;</span><br><span class="line">            printf(&quot;%d\n&quot;,ans[i]);</span><br><span class="line">        &#125;</span><br><span class="line">    &#125;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure>
<h2 id="总结"><a href="#总结" class="headerlink" title="总结"></a>总结</h2><p>完美的考察对于并查集，$tarjan$的理解，对于扩展域的使用在这里像极了$dp$，也让我发现自己对于tarjan的不足。</p>

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